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STatiana [176]
3 years ago
12

Yash wants to prove that KA=KC. To do that, he decides to prove that△KAB≅△KCB. Which theorem or postulate can Yash use to prove

the triangles congruent?
Hypotenuse-Leg (HL) Congruence TheoremSide-Side-Side (SSS) Congruence PostulateAngle-Angle-Side (AAS) Congruence TheoremSide-Angle-Side (SAS) Congruence Postulate

Mathematics
1 answer:
marta [7]3 years ago
6 0

Observe the figure clearly.

To Prove: KA = KC

Proof:

Consider the triangles \Delta KAB , \Delta KCB

KB = KB (Common side)

\angle KAB = \angle KCB (each angle is of 90 degree)

\angle ABK = \angle KBC (As BY bisects angle ABC)

Therefore, \Delta KAB \cong \Delta KCB

By AAS congruence Theorem which states:

"If two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent".

Hence, the given triangles are congruent by AAS congruence criteria.

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3 years ago
If f(x) = 3x2 − 8x, 0 ≤ x ≤ 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints.
denis23 [38]

Split up the interval [0, 3] into 6 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2], [2, 5/2], [5/2, 3]

The right endpoints are given by the arithmetic sequence,

r_i=0+\dfrac i2=\dfrac i2

with 1\le i\le6.

We approximate the integral of f(x) on the interval [0, 3] by the Riemann sum,

\displaystyle\int_0^3f(x)\,\mathrm dx=\sum_{i=1}^6f(r_i)\Delta x_i

\displaystyle=\frac{3-0}6\sum_{i=1}^6\left(3{r_i}^2-8r_i\right)

\displaystyle=\frac12\sum_{i=1}^6\left(\frac{3i^2}4-4i\right)

\displaystyle=\frac38\sum_{i=1}^6i^2-2\sum_{i=1}^6i

Recall the formulas,

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

Then the value of the integral is approximately

\displaystyle=\frac38\cdot\frac{6\cdot7\cdot13}6-2\cdot\frac{6\cdot7}2=\boxed{-\frac{63}8}=-7.875

Compare to the exact value of the integral, -9.

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3 years ago
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