Discussion
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- Question
- A point O in the interior of a rectangular ABCD is joined with each of the vertices A, B, C and D. Then :
Options- A. OA2 + OC2 = OB2 ? OD2
- B. OA2 + OC2 = OB2 + OD2
- C. OA2 = OB2 = OC2 + OD2
- D. OA2 + OD2 = OB2 + OC2
- Correct Answer
- OA2 + OC2 = OB2 + OD2
ExplanationAs per given figure , we can see that
Since the diagonals of a rectangle are equal and bisect each other. Let AC and BD intersect at M. Therefore M is the mid-point of AC and BD and AM = DM
From ?AOC, OA2 + OC2 = 2(AM2 + MO2) [Appollonius Theorem.] .......... ( 1 )
also in ?ODB, OB2 + OD2 = 2(MO2 + DM2) = 2(MO2 + AM2) .......... ( 2 )
From equations ( 1 ) and ( 2 ) .
? OA2 + OC2 = OB2 + OD2. - 1. If H is the orthocentre of ? ABC, then the orthocentre of ? HBC is (fig. given) :
Options- A. N
- B. M
- C. A
- D. L Discuss
- 2. In the given figure, In a ? ABC , ?B = ? C. If AM is the bisector of ? BAC and AN ? BC, then ? MAN is equal to :
Options- A. 1 (?B + ?C) 2
- B. 1 (?C- ?B ) 2
- C. ?B + ?C
- D. 1 (?B - ?C) 2 Discuss
- 3. The diagonals of a rectangle ABCD meet at 0. If ? BOC = 44°, then ? OAD is equal to :
Options- A. 90°
- B. 60°
- C. 100°
- D. 68° Discuss
- 4. ABCD is a rhombus with ? ABC = 56°, then ? ACD is equal to :
Options- A. 90°
- B. 60°
- C. 56°
- D. 62° Discuss
- 5. X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P; CX and BY are joined intersecting in Q. Then PXQY is a :
Options- A. Rectangle
- B. Rhombus
- C. Parallelogram
- D. Square Discuss
- 6. ABC is a ? in which AB = AC and D is a point on AC such that BC2 = AC × CD. Then :
Options- A. BD = DC
- B. BD = BC
- C. BD = AB
- D. BD = AD Discuss
- 7. The area of two similar ?s are 121 cm 2 and 81 cm 2 respectively. What is the ratio of their corresponding heights (altitudes):
Options- A. 11 9
- B. 22 9
- C. 11 18
- D. 33 6 Discuss
- 8. In ? ABC, the median BE intersects AC at E, if BG = 6 cm, where G is the centro-id, then BE is equal to :
Options- A. 7 cm
- B. 9 cm
- C. 8 cm
- D. 10 cm Discuss
- 9. If the sides of a triangle are produced then the sum of the exterior angles i.e., ? a + ? b + ? c is equal to :
Options- A. 180°
- B. 90°
- C. 360°
- D. 270° Discuss
- 10. In the given figure, side BC of ? ABC is produced to form ray BD and CE || BA. Then ? ACD is equal to :
Options- A. ?A - ?B
- B. 1 (?A + ?B) 2
- C. ?A + ?B
- D. 1 (?A - ?B) 2 Discuss
Plane Geometry problems
Search Results
Correct Answer: M
Explanation:
From above given triangle , It is clear that the altitudes AL, BM and CN of ?ABC intersect at H. Then H is the orthocentre of ?ABC.
In ?ABC, HL ? BC and BN ? CH.
Thus, the two altitudes HL and BN of ?HBC, intersects at A.
Hence the orthocentre of ?HBC is M .
Correct Answer: 1 (?B - ?C) 2
Explanation:
Correct Answer: 68°
Explanation:
Correct Answer: 62°
Explanation:
Correct Answer: Parallelogram
Explanation:
From above given figure ,
Proceeding as in Q. No. 4, we can prove that AXCY is a parallelogram .
Similarly, BXDY is a parallelogram.
Now, AXCY is a parallelogram .
? AY || CX
[? Opposite sides of a parallelogram are parallel]
? PY || QX ?(1)
Also, BXDY is a parallelogram
? DX || BY [? Opposite sides of a parallelogram are parallel]
? PX || QY ?(2)
Thus, in a quadrilateral PXQY,
From (i) and (ii) ,
we have , PY || QX and PX || QY
? PXQY is a parallelogram.
Correct Answer: BD = BC
Explanation:
Correct Answer: 11 9
Explanation:
Correct Answer: 9 cm
Explanation:
Correct Answer: 360°
Explanation:
From above figure , we have
Since every exterior angle is equal to sum of interior opposite angles,
So, ? a = A + B, ? b = B + C and ? c = A + C
We know that , A + B + C = 180°
? ?a +?b + ?c = 2(A + B + C) = 2 × 180° = 360°.
Correct Answer: ?A + ?B
Explanation:
From above given figure , we can see that
CE || BA and AC is the transversal.
? ?4 = ?1 (alternate interior ?s) .......... ( 1 )
Again, CE || BA and BD is the transversal.
? ?5 = ?2 (corresponding ?s) ................( 2 )
Adding equations ( 1 ) and ( 2 )
? ?4 + ?5 = ?1 + ?2
??ACD = ?A + ?B.
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