Hey There!
.
.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answers
Hey mate!
Thank you for asking this question! ❤
Answer:
___________________________________________________
Let ΔPQR and ΔABC be two similar triangles,
PQ/AB = QR/BC = PR/AC [Corresponding sides of similar triangles are in the same ratio] [1]
And as corresponding angles of similar triangles are equal
- ∠A = ∠P
- ∠B = ∠Q
- ∠C = ∠R
Construction: Draw PM ⏊ QR and AN ⏊ BC
In ΔPQR and ΔABC
- ∠PMR = ∠ANC [Both 90°]
- ∠R = ∠C [Shown above]
- ΔPQR ~ ΔABC [By Angle-Angle Similarity]
PM/AN = PR/AC [Corresponding sides of similar triangles are in the same ratio] [2]
Now, we know that
Area of a triangle
ar(ΔPQR)/ar(ΔABC) = PQ.PM/AB.AN
Therefore,
ar(ΔPQR)/ar(ΔABC) = (PQ/AB)² = (PR/AC)² = (QR/BC)²
[From 2 and 1]
[Hence:Proved] Q. E. D
____________________________________________________
<Judge It Yourself...>
Hope it helps you! ヅ
✪ Be Brainly ✪
Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.
we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
If ΔABC~ΔDEF
⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2}
ar(EFD)
ar(BCA)
=
DE
2
AB
2
→ (1)
As ΔABC~ΔDEF
\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}
DE
AB
=
EF
BC
=
2EQ
2BP
Hence, \frac{AB}{DE}=\frac{BP}{EQ}
DE
AB
=
EQ
BP
In ΔABP and ΔDEQ
\frac{AB}{DE}=\frac{BP}{EQ}
DE
AB
=
EQ
BP
∠B=∠E (∵ΔABC~ΔDEF)
By SAS rule, ΔABP~ΔDEQ
⇒ \frac{AB}{DE}=\frac{AP}{DQ}
DE
AB
=
DQ
AP
Squaring, we get
\frac{AB^2}{DE^2}=\frac{AP^2}{DQ^2}
DE
2
AB
2
=
DQ
2
AP
2
→ (2)
Comparing (1) and (2), we get
\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2}
ar(EFD)
ar(BCA)
=
DQ
2
AP
2
Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.