Mix Examples - Triangles Questions in English

(C) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $PR$ and $QS$ intersect at point $O$.
Since the diagonals bisect each other,$PO = OR = \frac{PR}{2} = \frac{9}{2} = 4.5$ and $QO = OS = \frac{QS}{2} = \frac{12}{2} = 6$.
In the right-angled triangle $\triangle POQ$,by the Pythagorean theorem:
$PQ^2 = PO^2 + QO^2$
$PQ^2 = (4.5)^2 + (6)^2$
$PQ^2 = 20.25 + 36 = 56.25$
$PQ = \sqrt{56.25} = 7.5$.
Since all sides of a rhombus are equal,the perimeter is $4 \times PQ$.
Perimeter $= 4 \times 7.5 = 30$.

(D) In a rhombus,the diagonals bisect each other at right angles. Let $AM = x$ and $DM = y$.
Since the diagonals intersect at $M$ at a $90^{\circ}$ angle,$\triangle AMB$ is a right-angled triangle.
By the Pythagorean theorem,$AM^2 + BM^2 = AB^2$.
Given $AM + DM = 17$,we have $x + y = 17$,so $y = 17 - x$.
In $\triangle AMB$,$x^2 + BM^2 = 13^2 = 169$.
In $\triangle AMD$,$x^2 + y^2 = AD^2$. Since all sides of a rhombus are equal,$AD = AB = 13$.
Thus,$x^2 + y^2 = 169$.
Substituting $y = 17 - x$: $x^2 + (17 - x)^2 = 169$.
$x^2 + 289 - 34x + x^2 = 169$.
$2x^2 - 34x + 120 = 0$.
$x^2 - 17x + 60 = 0$.
$(x - 12)(x - 5) = 0$.
So,$x = 12$ or $x = 5$.
If $x = 12$,then $y = 17 - 12 = 5$.
If $x = 5$,then $y = 17 - 5 = 12$.
Since $AC > BD$,$AM > DM$,so $x > y$. Thus,$x = 12$ and $y = 5$.
$DM = y = 5$.
Since the diagonals bisect each other,$BD = 2 \times DM = 2 \times 5 = 10$.

(A) Let $AB = x$ and $BC = y$.
In rectangle $ABCD$,$\triangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$.
By the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Given $x^2 + y^2 = 29^2 = 841$.
Also given $x + y = 41$.
We know that $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the values: $41^2 = 841 + 2xy$.
$1681 = 841 + 2xy$.
$2xy = 1681 - 841 = 840$.
$xy = 420$.
Now,we have $x + y = 41$ and $xy = 420$.
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$.
$t^2 - 41t + 420 = 0$.
Factoring the equation: $(t - 21)(t - 20) = 0$.
So,$t = 21$ or $t = 20$.
Since $AB > BC$,we have $AB = 21$ and $BC = 20$.
Thus,$AB = 21$.

(B) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.
Given $P = 30$, we have $3a = 30$, which implies $a = 10$.
The area of an equilateral triangle is given by the formula $Area = \frac{\sqrt{3}}{4} a^2$.
Substituting $a = 10$ into the formula:
$Area = \frac{\sqrt{3}}{4} (10)^2$
$Area = \frac{\sqrt{3}}{4} \times 100$
$Area = 25 \sqrt{3}$.

(A) Let $AB = x$.
Given $BC - AB = 4$,so $BC = x + 4$.
Given $AC - BC = 4$,so $AC = BC + 4 = (x + 4) + 4 = x + 8$.
Since $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$,by the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
Substituting the values: $x^2 + (x + 4)^2 = (x + 8)^2$.
$x^2 + x^2 + 8x + 16 = x^2 + 16x + 64$.
$x^2 - 8x - 48 = 0$.
$(x - 12)(x + 4) = 0$.
Since $x$ must be positive,$x = 12$.
Thus,$AB = 12$,$BC = 12 + 4 = 16$,and $AC = 16 + 4 = 20$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM \perp AC$.
By the geometric mean theorem in right triangles,$BM^2 = AM \cdot CM$.
Given $BM = 10$ and $CM = 5$,we have $10^2 = AM \cdot 5$,which gives $100 = 5 \cdot AM$,so $AM = 20$.
The hypotenuse $AC = AM + CM = 20 + 5 = 25$.
Using the Pythagorean theorem in $\Delta BMC$,$BC^2 = BM^2 + CM^2 = 10^2 + 5^2 = 100 + 25 = 125$,so $BC = \sqrt{125} = 5\sqrt{5}$.
Using the Pythagorean theorem in $\Delta AMB$,$AB^2 = BM^2 + AM^2 = 10^2 + 20^2 = 100 + 400 = 500$,so $AB = \sqrt{500} = 10\sqrt{5}$.
The perimeter of $\Delta ABC = AB + BC + AC = 10\sqrt{5} + 5\sqrt{5} + 25 = 25 + 15\sqrt{5}$.

(A) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC$. Let $AB = AC = x$ and $BC = y$.
The perimeter is $AB + AC + BC = 50$,so $2x + y = 50$,which implies $y = 50 - 2x$.
Since $\overline{AM}$ is an altitude to the base $\overline{BC}$ in an isosceles triangle,it bisects the base. Thus,$BM = MC = y/2 = (50 - 2x)/2 = 25 - x$.
In right-angled $\Delta ABM$,by the Pythagorean theorem: $AM^2 + BM^2 = AB^2$.
Substituting the values: $15^2 + (25 - x)^2 = x^2$.
$225 + 625 - 50x + x^2 = x^2$.
$850 - 50x = 0$,so $50x = 850$,which gives $x = 17$.
Now,find the base $y$: $y = 50 - 2(17) = 50 - 34 = 16$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AM$.
Area $= \frac{1}{2} \times 16 \times 15 = 8 \times 15 = 120$.

(B) In $\Delta ABC$,$AB = AC = 37$,which means $\Delta ABC$ is an isosceles triangle.
Since $\overline{AM}$ is the median to the base $BC$,it is also the altitude to the base $BC$ in an isosceles triangle.
Therefore,$\overline{AM} \perp \overline{BC}$ and $M$ is the midpoint of $BC$.
Since $BC = 70$,$BM = MC = \frac{70}{2} = 35$.
In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:
$AB^2 = AM^2 + BM^2$
$37^2 = AM^2 + 35^2$
$1369 = AM^2 + 1225$
$AM^2 = 1369 - 1225 = 144$
$AM = \sqrt{144} = 12$.

(C) Let $AB = x$.
Given $BC - AB = 7$,so $BC = x + 7$.
Given $AC - BC = 2$,so $AC = BC + 2 = (x + 7) + 2 = x + 9$.
Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$,by the Pythagorean theorem:
$AB^2 + BC^2 = AC^2$
$x^2 + (x + 7)^2 = (x + 9)^2$
$x^2 + x^2 + 14x + 49 = x^2 + 18x + 81$
$2x^2 + 14x + 49 = x^2 + 18x + 81$
$x^2 - 4x - 32 = 0$
$(x - 8)(x + 4) = 0$
Since $x$ represents a length,$x = 8$.
Thus,$AB = 8$,$BC = 8 + 7 = 15$,and $AC = 8 + 9 = 17$.
The perimeter of $\Delta ABC = AB + BC + AC = 8 + 15 + 17 = 40$.

(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.
Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.
Substituting the equal sides,we get: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 128$.
This simplifies to: $2(AB^{2} + BC^{2}) = 128$.
Dividing by $2$,we get: $AB^{2} + BC^{2} = 64$.
In the right-angled triangle $\triangle ABC$,by the Pythagorean theorem: $AC^{2} = AB^{2} + BC^{2}$.
Therefore,$AC^{2} = 64$.
Taking the square root of both sides,$AC = \sqrt{64} = 8$.

(A) Let the sides of the rectangle be $AB = x$ and $BC = y$.
The perimeter of a rectangle is given by $2(x + y) = 28$,which simplifies to $x + y = 14$.
In a rectangle,the diagonal $AC$ forms a right-angled triangle $ABC$ with sides $AB$ and $BC$.
By the Pythagorean theorem,$x^2 + y^2 = AC^2 = 10^2 = 100$.
We have the system of equations:
$1) x + y = 14$
$2) x^2 + y^2 = 100$
From $(1)$,$y = 14 - x$. Substituting this into $(2)$:
$x^2 + (14 - x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0$
$x^2 - 14x + 48 = 0$
$(x - 6)(x - 8) = 0$
So,$x = 6$ or $x = 8$.
If $x = 6$,then $y = 14 - 6 = 8$. Since $AB < BC$ $(6 < 8)$,this is a valid solution.
If $x = 8$,then $y = 14 - 8 = 6$. This contradicts $AB < BC$.
Therefore,the sides are $AB = 6$ and $BC = 8$.

(B) The area of an equilateral triangle with side length $a$ is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given that the area is $36 \sqrt{3}$,we can set up the equation:
$\frac{\sqrt{3}}{4} a^2 = 36 \sqrt{3}$.
Divide both sides by $\sqrt{3}$:
$\frac{1}{4} a^2 = 36$.
Multiply both sides by $4$:
$a^2 = 36 \times 4 = 144$.
Taking the square root of both sides:
$a = \sqrt{144} = 12$.
Therefore,the measure of its side is $12$.

(C) Let the two poles be $AB = 6 \, m$ and $CD = 11 \, m$ standing on a flat ground.
The distance between their bases is $BD = 12 \, m$.
Draw a line segment $AE$ parallel to $BD$ such that $E$ lies on $CD$.
Then $AE = BD = 12 \, m$ and $ED = AB = 6 \, m$.
Now,$CE = CD - ED = 11 \, m - 6 \, m = 5 \, m$.
In the right-angled triangle $\triangle AEC$,by the Pythagorean theorem:
$AC^2 = AE^2 + CE^2$
$AC^2 = (12)^2 + (5)^2$
$AC^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13 \, m$.
Thus,the distance between their tops is $13 \, m$.

(D) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC = 25$ and base $BC = 14$.
Since $\overline{AM}$ is an altitude to the base $BC$,in an isosceles triangle,the altitude from the vertex to the base also acts as the median.
Therefore,$M$ is the midpoint of $BC$.
Thus,$BM = MC = \frac{BC}{2} = \frac{14}{2} = 7$.
In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:
$AB^2 = AM^2 + BM^2$
$25^2 = AM^2 + 7^2$
$625 = AM^2 + 49$
$AM^2 = 625 - 49 = 576$
$AM = \sqrt{576} = 24$.
Hence,the length of the altitude $AM$ is $24$.

(A) Let the length of the ladder be $L$. The ladder, the wall, and the ground form a right-angled triangle.
According to the Pythagorean theorem, $L^2 = (\text{height})^2 + (\text{base})^2$.
In the first case, height $= 8 \, m$ and base $= 6 \, m$.
So, $L^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Thus, $L = \sqrt{100} = 10 \, m$.
In the second case, the same ladder of length $L = 10 \, m$ is used, and the new height is $6 \, m$.
Let the new base be $x$.
Using the Pythagorean theorem again: $L^2 = (\text{new height})^2 + x^2$.
$10^2 = 6^2 + x^2$.
$100 = 36 + x^2$.
$x^2 = 100 - 36 = 64$.
$x = \sqrt{64} = 8 \, m$.
Therefore, the distance of the lower end from the base of the wall is $8 \, m$.

(A) rhombus is a quadrilateral where all sides are equal and diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals be $d_1 = AC = 24$ and $d_2 = BD = 70$.
The diagonals bisect each other at point $O$. Thus,the segments are $AO = OC = \frac{24}{2} = 12$ and $BO = OD = \frac{70}{2} = 35$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 12^2 + 35^2$
$AB^2 = 144 + 1225 = 1369$
$AB = \sqrt{1369} = 37$.
The perimeter of the rhombus is $4 \times \text{side} = 4 \times 37 = 148$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
By the geometric mean theorem for the altitude: $BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = 4 \cdot 5 = 20$.
Therefore,$BM = \sqrt{20} = 2\sqrt{5}$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
Therefore,$AB = \sqrt{36} = 6$.
Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = CM^2 + BM^2 = 5^2 + (2\sqrt{5})^2 = 25 + 20 = 45$.
Therefore,$BC = \sqrt{45} = 3\sqrt{5}$.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QS \perp PR$.
By the property of geometric mean in a right-angled triangle,$PQ^2 = PS \cdot PR$.
Given $PQ = 6$ and $PS = 4$,we have $6^2 = 4 \cdot PR$,which implies $36 = 4 \cdot PR$,so $PR = 9$.
Since $PR = PS + RS$,we have $9 = 4 + RS$,so $RS = 5$.
Using the altitude theorem,$QS^2 = PS \cdot RS = 4 \cdot 5 = 20$,so $QS = \sqrt{20} = 2\sqrt{5}$.
In $\Delta QSR$,by the Pythagorean theorem,$QR^2 = QS^2 + RS^2 = 20 + 5^2 = 20 + 25 = 45$.
Thus,$QR = \sqrt{45} = 3\sqrt{5}$.

(A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.
Using the Pythagorean theorem,the hypotenuse $AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
The area of $\Delta ABC$ can be calculated in two ways:
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 7 \times 24 = 84$.
Also,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} \times AC \times BM = \frac{1}{2} \times 25 \times BM$.
Equating the two areas: $\frac{1}{2} \times 25 \times BM = 84$.
$25 \times BM = 168$.
$BM = \frac{168}{25} = 6.72$.

(B) From the diagram,there are $4$ steps in total.
Total horizontal distance (base) $= 4 \times 20 \, cm = 80 \, cm = 0.8 \, m$.
Total vertical distance (height) $= 4 \times 15 \, cm = 60 \, cm = 0.6 \, m$.
The distance between points $A$ and $B$ forms the hypotenuse of a right-angled triangle with base $0.8 \, m$ and height $0.6 \, m$.
Using the Pythagorean theorem: $AB = \sqrt{(0.8)^2 + (0.6)^2} = \sqrt{0.64 + 0.36} = \sqrt{1.00} = 1 \, m$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$. $N$ is a point on $AB$ and $M$ is a point on $BC$.
To prove: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Proof:
$1$. In right-angled $\Delta ABM$,by Pythagoras theorem: $AM^{2} = AB^{2} + BM^{2}$.
$2$. In right-angled $\Delta CBN$,by Pythagoras theorem: $CN^{2} = CB^{2} + BN^{2}$.
$3$. Adding these two equations: $AM^{2} + CN^{2} = AB^{2} + BM^{2} + CB^{2} + BN^{2}$.
$4$. Rearranging the terms: $AM^{2} + CN^{2} = (AB^{2} + BC^{2}) + (BM^{2} + BN^{2})$.
$5$. In $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
$6$. In right-angled $\Delta MBN$,by Pythagoras theorem: $MN^{2} = BM^{2} + BN^{2}$.
$7$. Substituting these into the equation from step $4$: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the statement is proved.

(A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BE \perp AC$.
By the property of similarity in right-angled triangles,if an altitude is drawn to the hypotenuse,the triangles on both sides of the altitude are similar to the whole triangle and to each other.
$1$. $\Delta ABE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{AB}{AC} = \frac{AE}{AB}$,which implies $AB^2 = AE \cdot AC$ --- $(1)$.
$2$. $\Delta CBE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{BC}{AC} = \frac{CE}{BC}$,which implies $BC^2 = CE \cdot AC$ --- $(2)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{AB^2}{BC^2} = \frac{AE \cdot AC}{CE \cdot AC}$.
Canceling $AC$ from the numerator and denominator,we get:
$\frac{AB^2}{BC^2} = \frac{AE}{CE}$.
Hence proved.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QD \perp PR$.
By the property of geometric mean in a right-angled triangle,we have $\Delta PDQ \sim \Delta QDR$.
From the similarity of triangles,the ratio of corresponding sides is equal: $\frac{PQ}{QR} = \frac{PD}{QD} = \frac{QD}{DR}$.
From $\frac{PQ}{QR} = \frac{QD}{DR}$,we get $PQ^2 = QR^2 \cdot \frac{PD}{DR}$.
Given $PD = 25 DR$,so $\frac{PD}{DR} = 25$.
Substituting this value,$PQ^2 = QR^2 \cdot 25$.
Taking the square root of both sides,we get $PQ = 5 QR$.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta QDR \sim \Delta PQR$.
$3$. From $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$4$. From the ratio $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = QD \cdot \frac{PQ}{QR}$.
$5$. From the ratio $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = RD \cdot \frac{PQ}{QR}$.
$6$. Substituting $QD$ in the expression for $PD$: $PD = (RD \cdot \frac{PQ}{QR}) \cdot \frac{PQ}{QR} = RD \cdot (\frac{PQ}{QR})^2$.
$7$. Given $PQ = 4QR$,so $\frac{PQ}{QR} = 4$.
$8$. Therefore,$PD = RD \cdot (4)^2 = 16RD$.

(N/A) Let the diagonals $XZ$ and $YW$ of the quadrilateral $\square XYZW$ intersect at point $O$ at right angles $(90^{\circ})$.
In $\triangle XOY$,$\triangle YOZ$,$\triangle ZOW$,and $\triangle WOX$,by the Pythagorean theorem:
$XY^{2} = XO^{2} + YO^{2}$ $(1)$
$YZ^{2} = YO^{2} + ZO^{2}$ $(2)$
$ZW^{2} = ZO^{2} + WO^{2}$ $(3)$
$XW^{2} = WO^{2} + XO^{2}$ $(4)$
Adding equations $(1)$ and $(3)$:
$XY^{2} + ZW^{2} = (XO^{2} + YO^{2}) + (ZO^{2} + WO^{2})$
Rearranging the terms:
$XY^{2} + ZW^{2} = (YO^{2} + ZO^{2}) + (WO^{2} + XO^{2})$
Substituting equations $(2)$ and $(4)$ into the expression:
$XY^{2} + ZW^{2} = YZ^{2} + XW^{2}$
Hence,it is proved.

(N/A) Let $ABCD$ be a rectangle with sides $AB = CD = l$ and $BC = DA = b$. Let the diagonals be $AC$ and $BD$.
In a rectangle,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle ABC$. By the Pythagoras theorem:
$AC^2 = AB^2 + BC^2 = l^2 + b^2$
Similarly,in the right-angled triangle $\triangle BCD$:
$BD^2 = BC^2 + CD^2 = b^2 + l^2$
Sum of the squares of the diagonals:
$AC^2 + BD^2 = (l^2 + b^2) + (b^2 + l^2) = 2l^2 + 2b^2$
Sum of the squares of the sides:
$AB^2 + BC^2 + CD^2 + DA^2 = l^2 + b^2 + l^2 + b^2 = 2l^2 + 2b^2$
Thus,the sum of the squares of the sides is equal to the sum of the squares of the diagonals.

(N/A) Given: In $\Delta PQR$,$\overline{QM} \perp \overline{PR}$. Let $PM = x$. Then $PQ = 2x$ and $RM = 3x$.
In right-angled $\Delta QMP$,by Pythagoras theorem: $PQ^2 = PM^2 + QM^2 \implies (2x)^2 = x^2 + QM^2 \implies 4x^2 = x^2 + QM^2 \implies QM^2 = 3x^2$.
In right-angled $\Delta QMR$,by Pythagoras theorem: $QR^2 = QM^2 + RM^2 \implies QR^2 = 3x^2 + (3x)^2 \implies QR^2 = 3x^2 + 9x^2 = 12x^2$.
Now,consider the sides of $\Delta PQR$: $PR = PM + RM = x + 3x = 4x$. So,$PR^2 = (4x)^2 = 16x^2$.
Check if $PQ^2 + QR^2 = PR^2$: $PQ^2 + QR^2 = (2x)^2 + 12x^2 = 4x^2 + 12x^2 = 16x^2$.
Since $PQ^2 + QR^2 = PR^2$,by the converse of Pythagoras theorem,$\Delta PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BM^2 = AM \cdot CM$.
Given that $BM = 2 CM$,substitute this into the equation:
$(2 CM)^2 = AM \cdot CM$
$4 CM^2 = AM \cdot CM$
Since $CM \neq 0$,we can divide both sides by $CM$ to get $AM = 4 CM$.
Now,the hypotenuse $AC$ is the sum of segments $AM$ and $CM$:
$AC = AM + CM$
$AC = 4 CM + CM$
$AC = 5 CM$.
Hence,the result is proved.

(N/A) Given: In $\Delta PQR$,$\overline{PM} \perp \overline{QR}$ and $PM^2 = QM \times RM$.
Step $1$: From the given equation,we have $\frac{PM}{QM} = \frac{RM}{PM}$.
Step $2$: In $\Delta PMQ$ and $\Delta RMP$,$\angle PMQ = \angle RMP = 90^\circ$ (since $\overline{PM}$ is an altitude).
Step $3$: By the $SAS$ similarity criterion,$\Delta PMQ \sim \Delta RMP$.
Step $4$: Since the triangles are similar,their corresponding angles are equal: $\angle QPM = \angle PRM$ and $\angle PQM = \angle RPM$.
Step $5$: In $\Delta PQR$,the sum of angles is $180^\circ$. Thus,$\angle P + \angle Q + \angle R = 180^\circ$.
Step $6$: Substituting the angles,we get $\angle QPM + \angle RPM + \angle Q + \angle R = 180^\circ$.
Step $7$: Since $\angle QPM = \angle R$ and $\angle RPM = \angle Q$,we have $\angle R + \angle Q + \angle Q + \angle R = 180^\circ$,which simplifies to $2(\angle Q + \angle R) = 180^\circ$,so $\angle Q + \angle R = 90^\circ$.
Step $8$: Therefore,$\angle P = 180^\circ - 90^\circ = 90^\circ$. Hence,$\Delta PQR$ is a right-angled triangle.

(N/A) $1$. In $\Delta ABC$,by the Pythagorean theorem,$AC^{2} = AB^{2} + BC^{2}$.
$2$. In right-angled $\Delta ABD$,$AD^{2} = AB^{2} + BD^{2}$. Since $D$ is the midpoint of $BC$,$BD = \frac{1}{2} BC$. Thus,$AD^{2} = AB^{2} + (\frac{1}{2} BC)^{2} = AB^{2} + \frac{1}{4} BC^{2}$.
$3$. In right-angled $\Delta CBF$,$CF^{2} = BC^{2} + BF^{2}$. Since $F$ is the midpoint of $AB$,$BF = \frac{1}{2} AB$. Thus,$CF^{2} = BC^{2} + (\frac{1}{2} AB)^{2} = BC^{2} + \frac{1}{4} AB^{2}$.
$4$. Adding the two equations: $AD^{2} + CF^{2} = (AB^{2} + \frac{1}{4} BC^{2}) + (BC^{2} + \frac{1}{4} AB^{2})$.
$5$. Simplifying: $AD^{2} + CF^{2} = (1 + \frac{1}{4}) AB^{2} + (1 + \frac{1}{4}) BC^{2} = \frac{5}{4} AB^{2} + \frac{5}{4} BC^{2}$.
$6$. Factoring out $\frac{5}{4}$: $AD^{2} + CF^{2} = \frac{5}{4} (AB^{2} + BC^{2})$.
$7$. Since $AB^{2} + BC^{2} = AC^{2}$,we get $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$. By Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.
$2$. Since $\overline{PM}$ is a median to side $\overline{QR}$,$M$ is the midpoint of $\overline{QR}$. Therefore,$QM = RM = \frac{1}{2} QR$,which implies $QR = 2RM$.
$3$. In right-angled $\Delta PQM$,by Pythagoras theorem,$PM^{2} = PQ^{2} + QM^{2}$. Thus,$PQ^{2} = PM^{2} - QM^{2}$.
$4$. Substitute $PQ^{2} = PM^{2} - QM^{2}$ and $QR = 2RM$ into the first equation: $PR^{2} = (PM^{2} - QM^{2}) + (2RM)^{2}$.
$5$. Since $QM = RM$,substitute $QM$ with $RM$: $PR^{2} = PM^{2} - RM^{2} + 4RM^{2}$.
$6$. Simplifying the expression,we get $PR^{2} = PM^{2} + 3RM^{2}$. Hence proved.

(N/A) Let $BC$ be divided into three equal parts by points $D$ and $E$. Let $BD = DE = EC = x$. Then $BE = 2x$ and $BC = 3x$.
In right-angled $\Delta ABE$,by Pythagoras theorem: $AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}$.
Multiplying by $8$: $8AE^{2} = 8AB^{2} + 32x^{2}$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2} = AB^{2} + 9x^{2}$.
Multiplying by $3$: $3AC^{2} = 3AB^{2} + 27x^{2}$.
Subtracting the two equations: $8AE^{2} - 3AC^{2} = (8AB^{2} + 32x^{2}) - (3AB^{2} + 27x^{2}) = 5AB^{2} + 5x^{2}$.
Since $DE = x$,$DE^{2} = x^{2}$. Thus,$8AE^{2} - 3AC^{2} = 5AB^{2} + 5DE^{2}$.

(N/A) Let the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$.
Since the diagonals of a rhombus bisect each other at right angles $(90^{\circ})$,we have $AO = OC = \frac{AC}{2}$ and $BO = OD = \frac{BD}{2}$.
In the right-angled triangle $\triangle AOB$,by the Pythagoras theorem:
$AB^{2} = AO^{2} + BO^{2}$
Substitute the values of $AO$ and $BO$:
$AB^{2} = (\frac{AC}{2})^{2} + (\frac{BD}{2})^{2}$
$AB^{2} = \frac{AC^{2}}{4} + \frac{BD^{2}}{4}$
$AB^{2} = \frac{AC^{2} + BD^{2}}{4}$
Multiplying both sides by $4$,we get:
$4AB^{2} = AC^{2} + BD^{2}$
Hence,it is proved that $AC^{2} + BD^{2} = 4AB^{2}$.

(A) $1$. In $\Delta ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$ (Equation $1$).
$2$. Since $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$. Thus,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
$3$. In $\Delta ABD$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AD^{2} = AB^{2} + BD^{2}$.
$4$. Rearranging for $BD^{2}$,we get $BD^{2} = AD^{2} - AB^{2}$.
$5$. Substitute $BC = 2BD$ into Equation $1$: $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$.
$6$. Substitute $BD^{2} = AD^{2} - AB^{2}$ into the expression: $AC^{2} = AB^{2} + 4(AD^{2} - AB^{2})$.
$7$. Simplify: $AC^{2} = AB^{2} + 4AD^{2} - 4AB^{2} = 4AD^{2} - 3AB^{2}$.
$8$. Hence,$AC^{2} = 4AD^{2} - 3AB^{2}$ is proved.

(N/A) In $\Delta ABC$,$\overline{AM}$ is the altitude to side $\overline{BC}$,so $\angle AMB = 90^{\circ}$ and $\angle AMC = 90^{\circ}$.
In $\Delta AMC$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$.
Since $M$ lies on $BC$,we have $MC = BC - BM$.
Substituting this into the equation: $AC^{2} = AM^{2} + (BC - BM)^{2}$.
Expanding the square: $AC^{2} = AM^{2} + BC^{2} + BM^{2} - 2 \cdot BC \cdot BM$.
In $\Delta AMB$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$.
Substituting $AM^{2} + BM^{2} = AB^{2}$ into the equation: $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.
Hence,the result is proved.

(N/A) Given: In quadrilateral $ABCD$,$\angle B = 90^{\circ}$ and $AD^{2} = AB^{2} + BC^{2} + CD^{2}$.
To prove: $\angle ACD = 90^{\circ}$.
Proof: Join $AC$. In $\triangle ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$.
Substitute this into the given equation: $AD^{2} = (AB^{2} + BC^{2}) + CD^{2}$.
This becomes $AD^{2} = AC^{2} + CD^{2}$.
In $\triangle ACD$,we have $AD^{2} = AC^{2} + CD^{2}$.
By the converse of the Pythagoras theorem,if the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite to the first side is a right angle.
Therefore,$\angle ACD = 90^{\circ}$.

(N/A) In $\Delta AMC$,$\angle AMC = 90^{\circ}$. By Pythagoras theorem,$AC^{2} = AM^{2} + MC^{2}$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.

(D) Let the starting point $A$ be at the origin $(0, 0)$ in a Cartesian coordinate system.
Moving $800 \, m$ north takes the person to point $B(0, 800)$.
Moving $500 \, m$ east from $B$ takes the person to point $C(500, 800)$.
Moving $400 \, m$ north from $C$ takes the person to point $D(500, 800 + 400) = D(500, 1200)$.
The direct distance from $A(0, 0)$ to $D(500, 1200)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$d = \sqrt{(500 - 0)^2 + (1200 - 0)^2} = \sqrt{500^2 + 1200^2}$.
$d = \sqrt{250000 + 1440000} = \sqrt{1690000}$.
$d = 1300 \, m$.

(A) Let the airport be at the origin $(0, 0)$.
After $t = 1.5 \, \text{hours}$,the distance covered by the first plane moving North is $d_1 = 800 \times 1.5 = 1200 \, km$.
The distance covered by the second plane moving East is $d_2 = 840 \times 1.5 = 1260 \, km$.
Since North and East directions are perpendicular,the distance between the two planes forms the hypotenuse of a right-angled triangle.
Using the Pythagorean theorem,the distance $D = \sqrt{d_1^2 + d_2^2}$.
$D = \sqrt{1200^2 + 1260^2} = \sqrt{1440000 + 1587600} = \sqrt{3027600}$.
$D = 1740 \, km$.

(N/A) Let the rectangle be $ABCD$ in the Cartesian plane. Let $A = (0, b)$,$B = (a, b)$,$C = (a, 0)$,and $D = (0, 0)$.
Let the coordinates of point $P$ be $(x, y)$.
Using the distance formula,the squared distances are:
$PA^2 = (x - 0)^2 + (y - b)^2 = x^2 + y^2 - 2by + b^2$
$PC^2 = (x - a)^2 + (y - 0)^2 = x^2 - 2ax + a^2 + y^2$
$PB^2 = (x - a)^2 + (y - b)^2 = x^2 - 2ax + a^2 + y^2 - 2by + b^2$
$PD^2 = (x - 0)^2 + (y - 0)^2 = x^2 + y^2$
Now,calculate the sums:
$PA^2 + PC^2 = (x^2 + y^2 - 2by + b^2) + (x^2 - 2ax + a^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
$PB^2 + PD^2 = (x^2 - 2ax + a^2 + y^2 - 2by + b^2) + (x^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
Since both sums are equal,$PA^2 + PC^2 = PB^2 + PD^2$ is proved.

(N/A) In $\Delta ABM$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$ $(1)$.
In $\Delta ACM$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$ $(2)$.
Subtracting $(2)$ from $(1)$: $AB^{2} - AC^{2} = BM^{2} - MC^{2} = (BM - MC)(BM + MC)$.
Since $D$ is the midpoint of $BC$,$BD = DC = \frac{BC}{2}$.
We can express $BM$ as $BD + DM$ and $MC$ as $DC - DM = BD - DM$.
Substituting these: $BM - MC = (BD + DM) - (BD - DM) = 2DM$.
Also,$BM + MC = BC$.
Therefore,$AB^{2} - AC^{2} = (2DM)(BC) = 2 \cdot BC \cdot DM$.

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
Given $AC = 5 CD$. Since $AC = AD + CD$,we have $AD + CD = 5 CD$,which implies $AD = 4 CD$.
Substituting $AD = 4 CD$ into the geometric mean equation:
$BD^2 = (4 CD) \cdot CD$
$BD^2 = 4 CD^2$
Taking the square root of both sides:
$BD = \sqrt{4 CD^2} = 2 CD$.
Thus,it is proved that $BD = 2 CD$.

(A) Given that in $\Delta ABC$,$m\angle A + m\angle C = m\angle B$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $m\angle A + m\angle B + m\angle C = 180^{\circ}$.
Substituting $m\angle A + m\angle C = m\angle B$ into the equation,we get $m\angle B + m\angle B = 180^{\circ}$,which implies $2m\angle B = 180^{\circ}$,so $m\angle B = 90^{\circ}$.
Since $\angle B = 90^{\circ}$,$\Delta ABC$ is a right-angled triangle with $AC$ as the hypotenuse.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the given values,$AC^2 = 7^2 + 24^2 = 49 + 576 = 625$.
Therefore,$AC = \sqrt{625} = 25$.

(B) Let the side of the square be $a$ and the diagonal be $d$.
The formula for the diagonal of a square is $d = a\sqrt{2}$.
Given $d = 10$,we have $a\sqrt{2} = 10$.
Therefore,$a = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.
The area of a square is given by $Area = a^2$.
$Area = (5\sqrt{2})^2 = 25 \times 2 = 50$.
Alternatively,the area of a square can be calculated using the diagonal as $Area = \frac{1}{2} \times d^2$.
$Area = \frac{1}{2} \times (10)^2 = \frac{1}{2} \times 100 = 50$.

(C) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta AMB \sim \Delta BMC$.
From the similarity $\Delta AMB \sim \Delta ABC$,we have the property: $BM^2 = AM \cdot MC$.
Given $AC = 25$ and $AM = 16$,we find $MC = AC - AM = 25 - 16 = 9$.
Substituting these values into the formula: $BM^2 = 16 \cdot 9$.
$BM^2 = 144$.
Taking the square root,$BM = \sqrt{144} = 12$.

(D) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PM} \perp \overline{QR}$.
According to the geometric mean theorem (or property of altitude to the hypotenuse),$PQ^2 = QM \cdot QR$.
Given $PQ = \sqrt{20}$,so $PQ^2 = 20$.
Substituting the values: $20 = 4 \cdot QR$.
Therefore,$QR = \frac{20}{4} = 5$.
Since $QR = QM + RM$,we have $5 = 4 + RM$.
Thus,$RM = 5 - 4 = 1$.

(A) Let the rhombus be $ABCD$ with diagonals $AC = 24$ and $BD = 10$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the intersection point be $O$.
Then,$AO = AC / 2 = 24 / 2 = 12$ and $BO = BD / 2 = 10 / 2 = 5$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 12^2 + 5^2$
$AB^2 = 144 + 25 = 169$
$AB = \sqrt{169} = 13$.
Since all sides of a rhombus are equal,the length of each side is $13$.

(B) In $\Delta XYZ$,since $m\angle Y = 90^{\circ}$,the triangle is a right-angled triangle.
Using the Pythagorean theorem,$XZ^2 = XY^2 + YZ^2$.
Substituting the given values,$XZ^2 = 3^2 + 4^2 = 9 + 16 = 25$.
Therefore,$XZ = \sqrt{25} = 5$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Since $M$ is the midpoint of $\overline{XZ}$,$\overline{YM}$ is the median to the hypotenuse $\overline{XZ}$.
Thus,$YM = \frac{1}{2} \times XZ = \frac{1}{2} \times 5 = 2.5$.

(C) Let the length of the ladder be $L = 17 \, m$.
Let the distance of the base of the ladder from the wall be $b = 8 \, m$.
Let the height reached by the ladder on the wall be $h$.
The wall,the ground,and the ladder form a right-angled triangle where the ladder is the hypotenuse.
According to the Pythagorean theorem: $h^2 + b^2 = L^2$.
Substituting the values: $h^2 + 8^2 = 17^2$.
$h^2 + 64 = 289$.
$h^2 = 289 - 64 = 225$.
$h = \sqrt{225} = 15 \, m$.
Thus,the upper end of the ladder reaches a height of $15 \, m$ on the wall.

(D) triangle is a right-angled triangle if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras theorem: $a^2 + b^2 = c^2$).
For option $A$: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. This is a right-angled triangle.
For option $B$: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This is a right-angled triangle.
For option $C$: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$. This is a right-angled triangle.
For option $D$: $8^2 + 24^2 = 64 + 576 = 640$,while $26^2 = 676$. Since $640 \neq 676$,this is not a right-angled triangle.