price of 20% in the price of rice is forces a person to buy 4 kg of rice is less for price 840 find the reduced price per kg and the original price per kg
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price of 20% in the price of rice is forces a person to buy 4 kg of rice is less for price 840 find the reduced price per kg and the original price per kg
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Answer:
. Find the area of a trapezium whose parallel sides are 28.7 cm and 22.3 cm, and the distance between them is 16cm with figure . Find the area of a trapezium whose parallel sides are 28.7 cm and 22.3 cm, and the distance between them is 16cm with figure Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.. Find the area of a trapezium whose parallel sides are 28.7 cm and 22.3 cm, and the distance between them is 16cm with figure Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.. Find the area of a trapezium whose parallel sides are 28.7 cm and 22.3 cm, and the distance between them is 16cm with figure Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.Answer:
Given : AD ⊥ BC
2DB = 3CD
Step-by-step explanation:
ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.