Advertisements
Advertisements
Question
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?
Advertisements
Solution
\[\frac{\text{ Area of smaller triangle }}{\text{ Area of bigger triangle }} = \frac{2}{3}\]
\[ \Rightarrow \frac{\frac{1}{2} \times \text{Height of smaller triangle } \times \text{ Base of smaller triangle }}{\frac{1}{2} \times \text{ Height of bigger triangle } \times \text{ Base of bigger triangle }} = \frac{2}{3}\]
\[ \Rightarrow \frac{6}{\text{ Base of bigger triangle }} = \frac{2}{3}\]
\[\Rightarrow \text{ Base of bigger triangle } = \frac{3}{2} \times 6\]
\[ = 9\]
APPEARS IN
RELATED QUESTIONS
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`

In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`

In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.

- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?

In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.

If ΔXYZ ~ ΔPQR then `(XY)/(PQ) = (YZ)/(QR)` = ?
Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.
Ratio of corresponding sides of two similar triangles is 4 : 7, then find the ratio of their areas = ?
In fig. BD = 8, BC = 12, B-D-C, then `(A(ΔABC))/(A(ΔABD))` = ?

In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.

(i) `(A(ΔABD))/(A(ΔADC))`
(ii) `(A(ΔABD))/(A(ΔABC))`
(iii) `(A(ΔADC))/(A(ΔABC))`
