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Answered on 27 Jun Learn The Adventure Of Toto
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered 6 days ago Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered 6 days ago Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered 6 days ago Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Let ABCD be a parallelogram.
∴∠A=∠C and ∠B=∠D (Opposite angles)
Let ∠A=x0 and ∠B=4x50
Now, ∠A+∠B=1800
(Adjacent angles are supplementary)
⇒x+4x50=1800
⇒9x5=1800
⇒x=20×5
⇒x=1000
Now, ∠A=1000 and ∠B=45×1000=800
Hence, ∠A=∠C=1000
∠B=∠D=800
read lessAnswered on 01 Jul Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 03 Jul Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Given: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.
In square ABCD,
AB = BC = CD = DA
And, AK = BL = CM = DN
(All sides of a square are equal.) (Given)
So, AB - AK = BC - BL = CD - CM = DA - DN
⇒ KB = CL = DM = AN.......... (1)
In △NAKand△KBL
∠NAK=∠KBL=900 (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)
\(\triangle DNM≅ \triangle CML \)
\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and ∠DNM=∠CML (Cpctc )… 4)
ML = LK and ∠CML=∠BLK (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, ∠DNM=∠AKN=∠KLB=∠LMC
Now,
In △NAK
∠NAK=900
Let ∠AKN=x0
So, ∠DNK=900+x0 (Exterior angles equals sum of interior opposite angles.)
⇒∠DNM+∠MNK=900+x0
⇒x0+∠MNK=900+x0
⇒ ∠MNK=900
Similarly,
∠NKL=∠KLM=∠LMN=900 ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 01 Jul Learn Construction
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Step 1: Draw a line segment AB=13 cm.
Step 2: At A, construct an angle of 45∘ and at B construct an angle of 90∘.
Step 3: Bisect these angles. Let bisector of these angles intersect at point X. Join AX and BX.
Step 4: Draw perpendicular bisectors of AX and BX to intersect AB at Y and Z respectively. Join AB and AC.
Answered on 01 Jul Learn Construction
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Given: In ∆ABC
AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.
Steps of Construction :
1. Draw DE = 10 cm
2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12∘
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.
Answered on 03 Jul Learn Construction
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45° and bisect it to get ∠QPX.
3. Construct an angle of 60° and bisect it to get ∠PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ABC is the required triangle.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 03 Jul Learn Construction
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
It is given that AB: BC: CA = 3: 4: 5
3x + 4x + 5x = 12
12x = 12
x = 1
AB = 3 cm, BC = 4 cm and CA = 5 cm
Steps of construction:
(1) Draw a sufficiently long line segment using a ruler.
(2) Locate points A and B on it such that AB = 3 cm.
(3) With A as the centre and radius 5 cm, draw an arc.
(4) With B as the centre and radius 4 cm, draw another arc that cuts the previous arc at C.
(5) Join AC and BC.
Then, ABC is the required triangle.
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