In the given figure ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm and the distance between AB and DC is 4 cm. Then the value of x is
Answer
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Hint: In this question it is given that ABCD is a trapezium, where the length of the sides AB = 7 cm and AD = BC = 5 cm,and DC = x cm. From point A and B two perpendicular lines are drawn which meets the side DC at L and M respectively, where AL = BM = 4cm. So to find the solution we need to use Pythagorean theorem in the right angle triangle $$\triangle ADL$$ and $$\triangle BMC$$, by this we will find the length of DL and MC and after that we will find
x = DC = DL + LM + MC.
Complete step-by-step answer:
In the above figure ABML is forming a rectangle, and as we know that the opposite sides of a rectangle are equal in length.
Therefore LM = AB = 7 cm.
Now for $$\triangle ADL$$ we are going to use the Pythagorean theorem which states that “The summation of the square of the base and perpendicular is equal to the square of the hypotenuse”.
i.e, $$\left( \text{Perpendicular} \right)^{2} +\left( \text{Base} \right)^{2} =\left( \text{Hypotenuse} \right)^{2} $$
$$\Rightarrow \left( AL\right)^{2} +\left( DL\right)^{2} =\left( AD\right)^{2} $$
$$\Rightarrow 4^{2}+\left( DL\right)^{2} =5^{2}$$
$$\Rightarrow \left( DL\right)^{2} =5^{2}-4^{2}$$
$$\Rightarrow \left( DL\right)^{2} =25-16$$
$$\Rightarrow \left( DL\right)^{2} =9$$
$$\Rightarrow DL=\sqrt{9}$$
$$\Rightarrow DL=3$$
Therefore the length of DL = 3 cm.
Now similarly we can find the length of MC.
By Pythagorean theorem we can write,
$$\left( BM\right)^{2} +\left( MC\right)^{2} =\left( BC\right)^{2} $$
$$\Rightarrow 4^{2}+\left( MC\right)^{2} =5^{2}$$
$$\Rightarrow \left( MC\right)^{2} =25-16$$
$$\Rightarrow \left( MC\right)^{2} =9$$
$$\Rightarrow MC=\sqrt{9}$$
$$\Rightarrow MC=3$$
Therefore DL = MC = 3 cm.
Hence,
x = DL + LM + MC
= (3 + 7 + 3) cm
= 13 cm.
Note: While solving this type of question you need to know that for a trapezium 1 pair of opposite sides (AB and DC) are parallel and another pair of opposite sides (AD and BC) are equal and also the angles on either side of the bases are the same size/measure.
x = DC = DL + LM + MC.
Complete step-by-step answer:
In the above figure ABML is forming a rectangle, and as we know that the opposite sides of a rectangle are equal in length.
Therefore LM = AB = 7 cm.
Now for $$\triangle ADL$$ we are going to use the Pythagorean theorem which states that “The summation of the square of the base and perpendicular is equal to the square of the hypotenuse”.
i.e, $$\left( \text{Perpendicular} \right)^{2} +\left( \text{Base} \right)^{2} =\left( \text{Hypotenuse} \right)^{2} $$
$$\Rightarrow \left( AL\right)^{2} +\left( DL\right)^{2} =\left( AD\right)^{2} $$
$$\Rightarrow 4^{2}+\left( DL\right)^{2} =5^{2}$$
$$\Rightarrow \left( DL\right)^{2} =5^{2}-4^{2}$$
$$\Rightarrow \left( DL\right)^{2} =25-16$$
$$\Rightarrow \left( DL\right)^{2} =9$$
$$\Rightarrow DL=\sqrt{9}$$
$$\Rightarrow DL=3$$
Therefore the length of DL = 3 cm.
Now similarly we can find the length of MC.
By Pythagorean theorem we can write,
$$\left( BM\right)^{2} +\left( MC\right)^{2} =\left( BC\right)^{2} $$
$$\Rightarrow 4^{2}+\left( MC\right)^{2} =5^{2}$$
$$\Rightarrow \left( MC\right)^{2} =25-16$$
$$\Rightarrow \left( MC\right)^{2} =9$$
$$\Rightarrow MC=\sqrt{9}$$
$$\Rightarrow MC=3$$
Therefore DL = MC = 3 cm.
Hence,
x = DL + LM + MC
= (3 + 7 + 3) cm
= 13 cm.
Note: While solving this type of question you need to know that for a trapezium 1 pair of opposite sides (AB and DC) are parallel and another pair of opposite sides (AD and BC) are equal and also the angles on either side of the bases are the same size/measure.
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