Question

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​

Answer 1

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.