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CAT 2020 Slot 3 — QA Question 4

Basics of QuadrilateralsEasy

In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trapezium in sq.cm is

Answer & solution

Correct answer: 28

Answer: 28

Solution

Easy

Drop the parallel sides into view: BC⊥DCBC\perp DC is the height. Drop a perpendicular from DD to ABAB; the 45∘45^\circ angle at AA creates an isosceles right triangle that gives the extra length of ABAB.

D C B A E 5 4 4 45°
1

Set up the height and the dropped perpendicular. DEDE is drawn from DD perpendicular to ABAB, so DEBCDEBC is a rectangle and DE=BC=4DE=BC=4. Also EB=DC=5EB=DC=5.

DE=BC=4⇒ EB=DC=5\begin{aligned} &DE=BC=4\\ &\Rightarrow\ EB=DC=5 \end{aligned}
2

Use the 45∘45^\circ angle. In right triangle AEDAED, ∠DAE=45∘\angle DAE=45^\circ and ∠AED=90∘\angle AED=90^\circ, so it is isosceles with AE=DEAE=DE.

AE=DE=4(isosceles right triangle)\begin{aligned} &AE=DE=4 \quad\text{(isosceles right triangle)} \end{aligned}
3

Length of ABAB.

AB=EB+AE=5+4(from steps 1 and 2)⇒ AB=9\begin{aligned} &AB=EB+AE=5+4 \quad\text{(from steps 1 and 2)}\\ &\Rightarrow\ AB=9 \end{aligned}
4

Area of the trapezium. Parallel sides DC=5DC=5 and AB=9AB=9, height BC=4BC=4.

Area=12 (DC+AB)×BC⇒ Area=12 (5+9)×4⇒ Area=12×14×4⇒ Area=28\begin{aligned} &\text{Area}=\tfrac{1}{2}\,(DC+AB)\times BC\\ &\Rightarrow\ \text{Area}=\tfrac{1}{2}\,(5+9)\times4\\ &\Rightarrow\ \text{Area}=\tfrac{1}{2}\times14\times4\\ &\Rightarrow\ \text{Area}=28 \end{aligned}
28 sq.cm28\ \text{sq.cm}

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CAT 2020 Slot 3 QA Q4: In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC — Solution | TheCATExam