In the given figure, it is given that AB=7m and CD=24m, find the length of DE and also show that triangle ABE and triangle BDC are congruent.
(Answer: 31m)
HOW???
Answers
Answer:
- DE = 31 m
- ΔABE ≅ ΔDCB by the SSS criterion
Step-by-step explanation:
To find:
- Length of DE
- Prove that ΔBDC ≅ ΔABE
Method to find:
First, let us find the length of DB.
Since ∠CDB is a right angle, ΔCDB becomes a right-angled triangle.
We know that in right-angled triangles:
A² + B² = C²
(By Pythagoras theorum, where A and B are the sides of the triangle and C is the hypotenuse)
So,
24² + DB² = 25²
⇒ DB² = 25² - 24²
⇒ DB² = 49
⇒ DB = √49
⇒ DB = 7 m
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Now, let us find BE by using the same method.
7² + BE² = 25²
⇒ BE² = 25² - 7²
⇒ BE² = 576
⇒ BE = √576
⇒ BE = 24 m
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DB + BE = DE
⇒ 24 + 7 = DE
⇒ DE = 31 m
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CD = BE = 24 m
CB = AB = 25 m
DB = AE = 7 m
Hence, ΔABE ≅ ΔDCB
(by the SSS criterion, i.e, If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.)
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Hope it helps!
Answer:
Now, A/Q
In ∆ABE and ∆BDC,
CB=AB=25m
DB=AE=7m
CD=BE=24m
∆ABE ≈∆BDC [ by SSS axiom ]
hence proved
