Math Olympiad Test: Triangles - 4 - Class 10 MCQ

Let AB and CD be the heights of given poles, BC be the distance between their feet and AD be the distance between their tops
Since, BCDE is a rectangle
∴ EB = DC = 9 m and ED = BC = 12 m
AE = AB – EB = 14 m – 9 m = 5m
Now, ΔAED is a right angled triangle
∴ AD² = AE² + ED² [By Pythagoras theorem]
⇒ AD² = (5)² + (12)² = 169 ⇒ AD = 13 m
Hence, the distance between their tops is 13 m.

Let AB be the height of tree which casts a shadow AC = 18 feet and ED be the height of child which casts a shadow CD = 2 feet
Now, in ΔABC and ΔDEC, we have 
∠A = ∠D [Each 90°]
∠C = ∠C [Common angle]
∴ ΔABC ~ ΔDEC [By AA similarity]
⇒ AB/DE = AC/DC
⇒ AB/3 = 18/2
⇒ AB = 27 feet
So, the height of tree is 27 feet.

Let the point A represent the airport.
Let the plane-I fly towards North.
∴ Distance of the plane-I from the airport after 1(1/2) hours
= Speed × Time = 1000 × (1)1/2 km= 1500 km

Let the plane-II fly towards West.
∴ Distance of the plane-II from the airport after (1)1/2 hours
= 1200 × (1)1/2km = 1800 km
Now, in right ∆ABC, using Pythagoras theorem, we have
BC² = AB² + AC²
⇒ BC² = (1800)² + (1500)² = 5490000
⇒ BC = 
Thus, after (1)1/2 hours, the two planes will be 300 √61 km apart from each other.

In ΔADM and ΔABF
∠DAM = ∠BAF (Common)
∠DMA = ∠BFM (Corresponding angles)
∴ ΔADM ~ ΔABF (By AA similarity)

∴ BC = BF + FC = 2BF = 48 cm (∵ CF = BF)

Since ΔABD is a right angled triangle
∴ AD² = AB² + BD² [By using Pythagoras theorem]
⇒ AD² = AB² + (BC/2)² [∵ AD is median on BC]
⇒ AD² = AB² + (1/4) BC² ...(i)
Similarly in, ΔBCE
CE² = BC² + BE² ⇒ CE² = BC² + (AB/2)² [∵ EC is median on AB]
⇒ CE² = BC² + (1/4).AB² ...(ii)
Adding (i) and (ii), we get

(i) Since, CD || AB [Given]
∴ In ΔRCD and ΔRBA, we have
∠BAR = ∠RDC [Alternate interior angles]
∠ABR = ∠RCD
∴ ΔRCD ~ ΔRBA [By AA similarity]
 ... (i)

i.e., Distance between the parks through town is 8.75 m.
(ii) In right ΔCRD, we have
(CD)² = (CR)² + (RD)²
⇒ RD² = (1.4)² – (1.2)² = 0.52 ⇒ RD = 0.72 m
Since ΔRCD ~ ΔRBA


i.e., Distance from Park A to Park B through point R = AR + RB
= 4.5 m + 7.5 m = 12 m

BPT (Basic proportional theorem) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the present ratio.

(P) Given : 
∴ ∠A is containing the sides AB and AC and ∠P is containing the sides PQ and PR.
∴ ΔABC ~ ΔPQR (By SAS criteria)
(Q) Given : ∠A = ∠P, ∠B = ∠Q
∴ ΔABC ~ ΔPQR (By AA criteria)

∵ Sides of the ΔABC and ΔPQR are in proportion
∴ ΔABC ~ ΔPQR (By SSS criteria)
(S) Given, DE || BC

(i) Since, XY || AC, we have
∠A = ∠BXY and ∠C = ∠BYX [Corresponding angles]
∴ ΔABC ~ ΔXBY [By AA similarity]
 ...(i)
But, ar(ΔABC) = 2 × ar(ΔXBY) [Given]
 ...(ii)
From (i) and (ii), we get 


Hence, AX : AB = (2 − √2) : 2
(ii) Since, ΔABC ~ ΔXBY

 

Construction: Join PQ, BP and AQ.
(i) 4AC² + BC² = 4AC² + (2QC)²
[∵ Q is mid-point of BC]
= 4AC² + 4QC²
= 4(AC² + QC²)
= 4AQ² [∵ AQC is right angled triangle]
(ii) 4BC² + AC² = 4BC² + (2CP)² [∵ P is mid-point of AC]
= 4BC² + 4CP² = 4(BC² + CP²) = 4BP2 [∵ PBC is right angled triangle]
(iii) 4[AC² + QC²] = 4(AQ)² [from (i) part]
⇒ 4AC² + BC² = 4AQ² ...(i)
4[BC² + CP²] = 4(BP)² [from (ii) part]
⇒ 4BC² + AC² = 4BP² ...(ii)
Now, adding (i) and (ii), we get
4(AQ² + BP²) = 5(AC² + BC²) = 5AB² [∵ ABC is right angled triangle]