In Fig. 15.78, ABCD is a trapezium in which AB=7 cm, AD=BC=5 cm, DC=x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Answers
Given : ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm.
Draw AL and BM are perpendiculars on DC, then
AL = BM = 4 cm and LM = 7 cm.
In right ΔBMC :
By Using Pythagoras theorem, we get
BC² = BM² + MC²
25 = 16 + MC²
MC² = 25 – 16
MC = √9
MC = 3 cm
Similarly ,
In right Δ ADL :
By Using Pythagoras theorem, we get
AD² = AL² + DL²
25 = 16 + DL²
DL² = 25 – 16
DL² = √9
DL = 3
Therefore, x = DC = (LD + ML + CM) = (3 + 7 + 3) = 13 cm
=> x = 13 cm
Now,
Area of trapezium ABCD ,A = ½ (AB + CD) × AL
A = ½ (7 + 13) × 4
A = 20 × 2
A = 40 cm²
Hence, Area of trapezium ABCD is 40 cm².
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Answer:-
Given :
ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm.
Draw AL and BM are perpendiculars on DC, then
AL = BM = 4 cm and LM = 7 cm.
In right ΔBMC :
By Using Pythagoras theorem, we get
BC² = BM² + MC²
25 = 16 + MC²
MC² = 25 – 16
MC = √9
MC = 3 cm
Similarly ,
In right Δ ADL :
By Using Pythagoras theorem, we get
AD² = AL² + DL²
25 = 16 + DL²
DL² = 25 – 16
DL² = √9
DL = 3
Therefore, x = DC = (LD + ML + CM) = (3 + 7 + 3) = 13 cm
=> x = 13 cm
Now,
Area of trapezium ABCD ,A = ½ (AB + CD) × AL
A = ½ (7 + 13) × 4
A = 20 × 2
A = 40 cm²
Hence, Area of trapezium ABCD is 40 cm².
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