In - ABC and - APQ - AP -5 cm and PB -10 cm- Find the ratio of the areas of - ABC and - APQ-A- 1- 9B- 3- 1C- 1- 3D- 9- 1

The correct option is D 9 : 1In Δ ABC and Δ APQ, ∠ PAQ is common and nbsp; ∠ APQ = ∠ ABC (corresponding angles) Δ ABC ∼Δ APQ (AA criterion) Since both triang ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Ratio of Area of Similar Triangles

In Δ ABC and ...

Question

In $Δ A B C$ and $Δ A P Q$ , AP = 5 cm and PB = 10 cm. Find the ratio of the areas of $Δ A B C a n d Δ A P Q .$

1 : 9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9 : 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 9 : 1
In $Δ A B C a n d Δ A P Q$ ,
$∠ P A Q$ is common and
$∠ A P Q = ∠ A B C$ (corresponding angles)
$Δ A B C \sim Δ A P Q$ (AA criterion)
Since both triangles are similar, using the theorem of areas of similar triangle we have
$\frac{a r e a o f Δ A B C}{a r e a o f Δ A P Q} = {(\frac{A B}{A P})}^{2} = {(\frac{15}{5})}^{2} = 9 : 1$

Suggest Corrections

Similar questions

Q. In below figure

P Q ∥ B C

and

A P : P B = 1 : 2

. Then the ratio of area

Δ A P Q

and

Δ A B C

will be:

Q. Question 8
If

Δ A B C \sim Δ P Q R, a n d \frac{B C}{Q R} = \frac{1}{3}, t h e n \frac{a r (Δ P R Q)}{a r (Δ B C A)}

is equal to

(A) 9
(B) 3
(C)

\frac{1}{3}

(D)

\frac{1}{9}

Q. In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of

Δ A P Q

\frac{1}{16}

of the area of

Δ A B C

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of $Δ A P Q$ is $\frac{1}{16}$ of the area of $Δ A B C$ .

In $Δ A B C$ , P divides the side AB such that AP : PB =1 : 2. Q is a point in AC sucht that PQ||BC. Find the ratio of the areas of $Δ A P Q$ and trapezium BPQC.

Join BYJU'S Learning Program

In ΔABC and ΔAPQ, AP = 5 cm and PB = 10 cm. Find the ratio of the areas of ΔABC and ΔAPQ.

The correct option is C 9 : 1In ΔABC and Δ APQ, ∠PAQ is common and ∠APQ=∠ABC (corresponding angles) ΔABC∼ΔAPQ (AA criterion) Since both triangles are similar, using the theorem of areas of similar triangle we have area of ΔABCareaofΔAPQ=(ABAP)2=(155)2=9:1

In $Δ A B C$ and $Δ A P Q$ , AP = 5 cm and PB = 10 cm. Find the ratio of the areas of $Δ A B C a n d Δ A P Q .$