Home
Class 10
MATHS
It is given that Delta ABC - Delta PQR, ...

It is given that `Delta ABC - Delta PQR`, with `(BC)/(QR) = (1)/(4)`. Then `(ar Delta PQR)/(ar Delta ABC)` is equal to:

A

16

B

3

C

`(1)/(3)`

D

`(1)/(9)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the ratio of the areas of triangles PQR and ABC given that the triangles are similar and the ratio of the lengths of one pair of corresponding sides is given. ### Step-by-Step Solution: 1. **Understand the Given Information:** We know that triangles ABC and PQR are similar. The ratio of the lengths of the sides BC and QR is given as: \[ \frac{BC}{QR} = \frac{1}{4} \] 2. **Use the Property of Similar Triangles:** For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore: \[ \frac{ar \Delta PQR}{ar \Delta ABC} = \left(\frac{QR}{BC}\right)^2 \] 3. **Substitute the Given Ratio:** We know that: \[ \frac{BC}{QR} = \frac{1}{4} \] This implies: \[ \frac{QR}{BC} = 4 \quad \text{(by taking the reciprocal)} \] 4. **Calculate the Area Ratio:** Now, substituting this into the area ratio formula: \[ \frac{ar \Delta PQR}{ar \Delta ABC} = \left(4\right)^2 = 16 \] 5. **Final Answer:** Thus, the ratio of the areas of triangle PQR to triangle ABC is: \[ \frac{ar \Delta PQR}{ar \Delta ABC} = 16 \]
Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    OSWAL PUBLICATION|Exercise ASSERTION AND REASON BASED MCQs|4 Videos
  • TRIANGLES

    OSWAL PUBLICATION|Exercise CASE - BASED MCQs|15 Videos
  • SURFACE AREAS AND VOLUMES

    OSWAL PUBLICATION|Exercise BOARD CORNER (LONG ANSWER TYPE QUESTIONS)|9 Videos

Similar Questions

Explore conceptually related problems

It is given that Delta ABC~Delta PRQ and (BC)/(QR)=(2)/(3) " then " (ar (Delta PQR))/(ar (Delta ABC))=?

If triangle ABC - triangle PQR and (BC)/(QR) = 1/4 then (ar(trianglePQR))/(ar(triangleABC)) =

In the given figure Delta ABC - Delta PQR . Find the value of x.

In Delta ABC and Delta DEF , we have (AB)/(DE)=(BC)/(EF)=(AC)/(DF)=5/7 , then ar (Delta ABC) : ar (Delta DFR) =?

If Delta ABC ~ Delta PQR and 4A (Delta ABC) = 25 A (Delta PQR) then AB : PQ = ?

Given that : Delta ABC ~ DeltaPQR , If ("area " (Delta PQR))/("area " (Delta ABC)) = (256)/(441) and PR = 12 cm, then AC is equal to

Delta ABC ~ Delta PQR and ar(DeltaABC)=4ar(DeltaPQR) IF BC=12cm then find QR

The base BC of Delta ABC is divided at D, so that BD=(1)/(2)DC. Prove that ar(Delta ABD)=(1)/(3)ar(Delta ABC)

Given /_ABC=/_PQR if (AB)/(PQ)=(1)/(3) then find (ar/_ABC)/(ar/_PQR)

If PS is median of the triangle PQR, then ar( Delta PQS): ar( Delta QRP) is