Areas of two similar triangles are 36, cm^{2} | vedclass

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

Question

(D) Given,area of smaller triangle $= 36\, cm^{2}$ and area of larger triangle $= 100\, cm^{2}$.Also,length of a side of the larger triangle $= 20\, c

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Given,area of smaller triangle $= 36\, cm^{2}$ and area of larger triangle $= 100\, cm^{2}$.Also,length of a side of the larger triangle $= 20\, cm$.Let the length of the corresponding side of the smaller triangle be $x\, cm$.By the property of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.$\frac{	ext{Area of larger triangle}}{	ext{Area of smaller triangle}} = \frac{(	ext{Side of larger triangle})^{2}}{(	ext{Side of smaller triangle})^{2}}$$\Rightarrow \frac{100}{36} = \frac{(20)^{2}}{x^{2}}$$\Rightarrow x^{2} = \frac{20^{2} 	imes 36}{100} = \frac{400 	imes 36}{100} = 4 	imes 36 = 144$$x = \sqrt{144} = 12\, cm$.Hence,the length of the corresponding side of the smaller triangle is $12\, cm$.

Areas of two similar triangles are $36\, cm^{2}$ and $100\, cm^{2}$. If the length of a side of the larger triangle is $20\, cm$,find the length of the corresponding side of the smaller triangle. (in $cm$)

Similar Questions

State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

The areas of two similar triangles are $9$ and $16$. Then,the ratio of their corresponding sides is...............

In $\Delta PQR$ and $\Delta XYZ$,$\angle P \cong \angle X$ and $\angle Q \cong \angle Z$. If $PQ = 9$,$QR = 6$,$PR = 4.5$,and $XY = 7.5$,find $YZ$ and $XZ$.

Let $\Delta ABC \sim \Delta DEF$ and their areas be,respectively,$64 \text{ cm}^2$ and $121 \text{ cm}^2$. If $EF = 15.4 \text{ cm}$,find $BC$ (in $\text{cm}$).

In a $\triangle PQR$,$PR^{2} - PQ^{2} = QR^{2}$ and $M$ is a point on side $PR$ such that $QM \perp PR$. Prove that $QM^{2} = PM \times MR$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module