(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion (alternate interior angles are equal

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

(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion (alternate interior angles are equal).Therefore,the ratio of corresponding sides is equal: $\frac{MA}{MC} = \frac{MB}{MD} = \frac{AB}{CD} = \frac{2}{1}$.Given $AC = MA + MC = 15$.Substituting $MA = 2MC$ into the equation: $2MC + MC = 15$.$3MC = 15 \implies MC = 5$.Then,$MA = 2 \times 5 = 10$.Thus,$MA = 10$ and $MC = 5$.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $\frac{AB}{CD} = \frac{2}{1}$ and $AC = 15$,find $MA$ and $MC$.

A
$MA = 10, MC = 5$
B
$MA = 5, MC = 10$
C
$MA = 7.5, MC = 7.5$
D
$MA = 9, MC = 6$

Explore More

Browse All

Question Bank

All Subjects

Class 10 Questions

Class 10

Mathematics Questions

Chapter

Triangles

Subtopic

Mix Examples - Triangles

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,$2 AB = PQ$,and $BC = 10$,then $QR = \dots$

Medium

View Solution

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB = 3, BC = 4, AC = 5$ and $QR = 6$,find $PQ$ and $PR$.

Medium

View Solution

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$,$BC$,and $BM$.

Medium

View Solution

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

Medium

View Solution

$\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow FDE$. If $m \angle E + m \angle F = 130^{\circ}$,then $m \angle B = \dots$ (in $^{\circ}$)

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $\frac{AB}{CD} = \frac{2}{1}$ and $AC = 15$,find $MA$ and $MC$.

Similar Questions

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,$2 AB = PQ$,and $BC = 10$,then $QR = \dots$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB = 3, BC = 4, AC = 5$ and $QR = 6$,find $PQ$ and $PR$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$,$BC$,and $BM$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

$\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow FDE$. If $m \angle E + m \angle F = 130^{\circ}$,then $m \angle B = \dots$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module