The ratios of the sides of a triangle ABC are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the sides of the triangle be $5x$,$12x$,and $13x$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.The area

Vedclass · Accepted Answer

$15, 36, 39$

Vedclass · Answer

(C) Let the sides of the triangle be $5x$,$12x$,and $13x$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.The area of a right-angled triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Area $= \frac{1}{2} 	imes 5x 	imes 12x = 30x^2$.Given that the area is $270$,we have $30x^2 = 270$.$x^2 = 9$,which implies $x = 3$.The sides are $5(3) = 15$,$12(3) = 36$,and $13(3) = 39$.Thus,the sides are $15, 36, 39$.

The ratios of the sides of a triangle $ABC$ are $5:12:13$ and its area is $270$. Then the sides of the triangle are:

Similar Questions

In $\triangle ABC$,if $AB: BC: CA = 6: 4: 5$,then $R: r =$

In $\Delta ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} =$

In a triangle $ABC$,if $a \neq b$,then the value of $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$ is:

If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then

The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module