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 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 ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.

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

In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.

`(A(triangleABD))/(A(triangleABC))`
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.
if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.

If ΔXYZ ~ ΔPQR then `(XY)/(PQ) = (YZ)/(QR)` = ?
Ratio of corresponding sides of two similar triangles is 4 : 7, then find the ratio of their areas = ?
In fig., PM = 10 cm, A(ΔPQS) = 100 sq.cm, A(ΔQRS) = 110 sq.cm, then NR?

ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding `square`.
`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
`100/110 = (square)/(NR)`,
NR = `square` cm
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?

From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `(A(ΔABC))/(A(ΔBCD))` = ?

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))`
Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
- Draw two triangles, give the names of all points, and show heights.
- Write 'Given' and 'To prove' from the figure drawn.
If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
