Question

Q.17 In A ABC, MN || BC, the area of quadrilateral MBCN=130 sqcm. If AN : NC=4: 5, then the area of triangleMAN is:

Ans
X 1. 45 cm?
X 2.65 cm
✓ 3. 32 cm
X 4. 40 cm?​

Answer 1

In A ABC, MN || BC, the area of quadrilateral MBCN=130 sqcm. If AN : NC=4: 5, then the area of triangle MAN is: 32 cm^2

Option 3. is correct.

Given,

MN || BC

The area of quadrilateral MBCN=130 sqcm

AN : NC=4: 5

⇒ AC = 9 cm

From given, we have,  Δ ABC ~ Δ AMN,

∴ (ar Δ ABC) / (ar Δ AMN) = (AB / AN)^2

(ar Δ ABC) / (ar Δ AMN) = (9 / 4)^2

∴  (ar Δ ABC) / (ar Δ AMN) = 81 / 16

⇒ ar Δ ABC = 81/16 × ar Δ AMN ..........(1)

Given, triangle ABC is the sum of triangle AMN and quadrilateral MBCN, we have,

ar Δ ABC = ar Δ AMN + ar quad MBCN

using (1) and given area of quadrilateral, we have,

Using "x" as the area of Δ AMN

130 + x = 81/16 x

130 = 81/16 x - x

130 = 65/16 x

x = 130 × 16/65 = 32

∴ (ar ΔAMN) = 32 sq cm.

Answer 2

Area of ΔMAN is 32 cm². Means option third is correct.

Step-by-step explanation:

• Here it is given that

        $\mathbf{\frac{AN}{NC}=\frac{4}{5}}$

        So

        $\mathbf{\frac{AN}{AB}=\frac{4}{9}}$        ...1)

• Here it is also given that

        $\mathbf{MN\parallel BC}$

        So

        $\mathbf{\Delta MAN\sim \Delta ABC}$

• It is also clear that

        $\mathbf{ar\Delta ABC-ar\Delta MAN =area \ of \trapezium \ MBCN = 130 \ cm^{2}}$        ...2)

• From theorem of ratio of similar triangle is ratio of square of corresponding side.

        We can write in this way

        $\mathbf{\frac{ar\Delta ABC}{ar\Delta MAN}=\left ( \frac{AB}{AN} \right )^{2}}$

        $\mathbf{\frac{ar\Delta ABC}{ar\Delta MAN}=\left ( \frac{9}{4} \right )^{2}}$

        $\mathbf{\frac{ar\Delta ABC}{ar\Delta MAN}=\frac{81}{16} }$            ...3)

• Subtract 1 on both side of equation 1), we get

        $\mathbf{\frac{ar\Delta ABC}{ar\Delta MAN}-1=\frac{81}{16} -1}$

        $\mathbf{\frac{ar\Delta ABC-ar\Delta MAN}{ar\Delta MAN}=\frac{81-16}{16}}$         ...4)

• From equation 2) and equation 4), Equation 4) can be written as

        $\mathbf{\frac{Area\ of\ MBCN}{ar\Delta MAN}=\frac{65}{16}}$

        So

       $\mathbf{\frac{130}{ar\Delta MAN}=\frac{65}{16}}$

       On solving above, we get

       Area of ΔMAN =32 cm²