Question

base of a triangle is 9 and height is 5 base of another triangle is 10 and height is 6 find the ratio of areas of these triangle

Answer 1

Heya !!

Here's your answer.. ⬇⬇

♦ Given :- Let The triangle be ∆ABC and ∆PQR.
In ∆ABC,
Base of ∆ABC is BC = 9cm
Altitude of ∆ABC is AE = 5cm

In ∆PQR,
Base is QR = 10cm
Altitude is PM = 6cm

♦ To Find :- Ratio of Area of ∆ABC and ∆PQR

♦ Solution :-

$= \frac{area \: \: of \: \: abc}{area \: \: of \: \: pqr} \\ \\ = \frac{ \frac{1}{2} \times ae \times bc }{ \frac{1}{2} \times pm \times qr } \\ \\ = \frac{ \frac{1}{2} \times 5 \times 9 }{ \frac{1}{2} \times 6 \times 10} \\ \\ = \frac{3}{4}$


Hence, Ratio of Area of ∆ABC to ∆PQR is 3:4



Hope it helps..

Thanks. (:

Answer 2

Answer:

The required ratio is 3 : 4.

Step-by-step explanation:

Let,

The triangle having 9 units as base and 5 units as height is ∆ ABC

And,the triangle having 10 units as base and 6 units as height is ∆ PQR

___________________

Now,

Area of ∆ ABC = (base × height)/2

Area of ∆ ABC =(9×5)/2

Area of ∆ ABC = 45/2 = 22.5 unit²

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Area of ∆ PQR = (10×6)/2

Area of ∆ PQR = 10×3=30 unit²

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Ratio = 22.5 : 30

Ratio= 225:300

Ratio = 3:4