NTSE Stage 2 SAT 2019 Exam Question Paper

Question : 55 of 100

Marks: +1, -0

Let D be a point on the side BC of a triangle ABC suchthat ∠ADC = ∠BAC. If AC = 21 cm, then the side of anequilateral triangle whose area is equal to the area of therectangle with sides BC and DC is

$14×3^1/2$
$42×3^{-1}/2$
$14×3^3/4$
$42×3^1/2$

Validate

Solution:

∵

∠BAC=∠ADC

(given)

∠C=∠C

(common)

∴ΔABC~ΔDAC

(by AA similarity criterion)

⇒\;\;{BC}/{AC}=\;\;{AC}/{DC}

⇒BC×DC=AC^{2}

⇒BC×DC=(21)^{2}

= area of rectangle with sides BC & DC
Now, Area of equilateral triangle = area of rectangle

⇒\;\;{√{3}}/{4}(\;\text"side"\;)^{2}=(21)^{2}⇒\;\text"Side"\;=14×3^{3/4}

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