IPMAT Indore 2023 (MCQ) - In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be | PYQs + Solutions

IPMAT Indore 2023

Geometry

Triangles

Medium

In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be

11:4\sqrt{10}

12:11

11:12

4\sqrt{10}:11

✅ Correct Option: 1

Let AB = 2k, BC = 4k, and CA = 3k, where k is some positive constant.

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Place B at the origin (0,0) and C at (4k,0) so BC = 4k along the x-axis.

Since D is the midpoint of BC, D = (2k,0).

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Let's find the coordinates of A. Since AB = 2k and AC = 3k, if A = (x,y) where y > 0, then:

From the Pythagorean theorem:

AB^2 = x^2 + y^2 = (2k)^2 = 4k^2

AC^2 = (x-4k)^2 + y^2 = (3k)^2 = 9k^2

From

AB^2 = 4k^2

x^2 + y^2 = 4k^2

From

AC^2 = 9k^2

(x-4k)^2 + y^2 = 9k^2

x^2 - 8kx + 16k^2 + y^2 = 9k^2

Substituting

y^2 = 4k^2 - x^2

x^2 - 8kx + 16k^2 + 4k^2 - x^2 = 9k^2

-8kx + 20k^2 = 9k^2

-8kx = -11k^2

x = \dfrac{-11k^2}{-8k} = \dfrac{11k}{8}

Now for y:

y^2 = 4k^2 - x^2 = 4k^2 - (\dfrac{11k}{8})^2

y^2 = 4k^2 - \dfrac{121k^2}{64}

y^2 = \dfrac{256k^2 - 121k^2}{64}

y^2 = \dfrac{135k^2}{64}

y = \dfrac{\sqrt{135}k}{8}

A = (\dfrac{11k}{8}, \dfrac{\sqrt{135}k}{8})

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Since AM is the altitude to BC, M lies on BC with coordinates M = (m,0), where AM ⊥ BC.

This means

(A-M) \cdot (C-B) = 0

(A-M) = (\dfrac{11k}{8} - m, \dfrac{\sqrt{135}k}{8})

(C-B) = (4k, 0)

Their dot product:

(\dfrac{11k}{8} - m)(4k) + (\dfrac{\sqrt{135}k}{8})(0) = 0

\dfrac{11k^2}{2} - 4km = 0

m = \dfrac{11k}{8}

M = (\dfrac{11k}{8}, 0)

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Calculating BM and AD:

BM = |M - B| = |(\dfrac{11k}{8}, 0) - (0, 0)| = \dfrac{11k}{8}

AD = |A - D| = |(\dfrac{11k}{8}, \dfrac{\sqrt{135}k}{8}) - (2k, 0)|

AD = \sqrt{(\dfrac{11k}{8} - 2k)^2 + (\dfrac{\sqrt{135}k}{8})^2}

AD = \sqrt{(\dfrac{11k - 16k}{8})^2 + (\dfrac{\sqrt{135}k}{8})^2}

AD = \sqrt{(\dfrac{-5k}{8})^2 + (\dfrac{\sqrt{135}k}{8})^2}

AD = \sqrt{\dfrac{25k^2}{64} + \dfrac{135k^2}{64}}

AD = \sqrt{\dfrac{160k^2}{64}}

AD = \dfrac{k\sqrt{160}}{8} = \dfrac{k\sqrt{16 \cdot 10}}{8} = \dfrac{4k\sqrt{10}}{8} = \dfrac{k\sqrt{10}}{2}

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The ratio of BM to AD:

BM : AD = \dfrac{11k}{8} : \dfrac{k\sqrt{10}}{2}

BM : AD = \dfrac{11k}{8} : \dfrac{k\sqrt{10}}{2} = \dfrac{11k}{8} \cdot \dfrac{2}{k\sqrt{10}} = \dfrac{11 \cdot 2}{8 \cdot<div style="height: 10px"></div>\sqrt{10}} = \dfrac{22}{8\sqrt{10}} = \dfrac{11}{4\sqrt{10}}

Therefore, the ratio of BM to AD is

\dfrac{11}{4\sqrt{10}}

\dfrac{11\sqrt{10}}{40}