Question

Base of triangles is 9 and hejaht 5. Base of another
triangle is 10 and helght is 6 Find the ration
of areas of these triangles.ans
​

Answer 1

GIVEN :-

• Base and height of 1st ∆ = 9 units and 5 units.
• Base and height of 2nd ∆ = 10 units and 6 units.

TO FIND :-

• Ratio of both triangles.

SOLUTION :-

Let the area of first ∆ be "A1" and the area of second triangle be "A2".

$\\ : \implies \displaystyle \sf \: A1: A2 = \frac{ \bigg(\dfrac{1}{2} \times base \times height \bigg)}{ \bigg(\dfrac{1}{2} \times base \times height \bigg)}$

$: \implies \displaystyle \sf \: A1: A2 = \frac{ \bigg( \dfrac{1}{2} \times 9 \times 5 \bigg)}{ \bigg(\dfrac{1}{2} \times10 \times 6 \bigg) }$

$: \implies \displaystyle \sf \: A1: A2 = \frac{\bigg( \dfrac{45}{2} \bigg)}{ \bigg(\dfrac{60}{2} \bigg) }$

$: \implies \displaystyle \sf \: A1: A2 = \dfrac{45}{2} \times \dfrac{2}{60}$

$: \implies \displaystyle \sf \: A1: A2 = \frac{45}{60}$

$: \implies \displaystyle \sf \: A1: A2 = \frac{3}{4}$

$: \implies \underline{ \boxed{\displaystyle \sf \: A1: A2 = 3:4}}$

Answer 2

$\rm\large{Answer :-}$

Given :-

$\mapsto$ Base of triangle is 9 and height is 5. Base of another triangle is 10 and height is 6.

To Find :-

$\mapsto$ What is the ratio of areas of these triangle.

Solution :-

Let, the triangle be ∆ABC and ∆PQR.

$\leadsto$ In ∆ABC,

• Base of ∆ABC is BC = 9cm
• Altitude of ∆ABC is AE = 5cm

$\leadsto$ In ∆PQR,

• Base of ∆PQR is QR = 10cm
• Altitude of ∆PQR is PM = 6cm

▪️According to the question :-

$\longrightarrow$ $\dfrac{Area(∆ABC)}{Area(PQR)}$

$\implies$ $\sf\dfrac{\dfrac{1}{2}×AC×BC}{\dfrac{1}{2}×PM×QR}$

$\implies$ $\sf\dfrac{\dfrac{1}{2}×5×9}{\dfrac{1}{2}×6×10}$

$\dashrightarrow$ $\dfrac{3}{4}$

$\therefore$ The Ratio of Area of ∆ABC : Area of ∆PQR = $\boxed{\bold{\large{3:4}}}$

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